Question

Successive numbers starting from `1` are written in a rectangular grid, starting in the top left corner and snaking down diagonally to the right as shown. In which row and column does `2008` occur?

`color(white)(0)1color(white)(00)2color(white)(00)6color(white)(00)7color(white)(0)15color(white)(0)16`
`color(white)(0)3color(white)(00)5color(white)(00)8color(white)(0)14color(white)(0)"etc"`
`color(white)(0)4color(white)(00)9color(white)(0)13`
`10color(white)(0)12`
`11`

Answer 1

The `9`th row, `55`th column.

Explanation:

The last number added to the `n`th (reverse) diagonal is the `n`th triangular number `T_n = 1/2n(n+1)`.

What is the smallest triangular number greater than or equal to `2008`?

Given:

`1/2n(n+1) = T_n >= 2008`

Multiply both ends by `2` to get:

`n(n+1) >= 4016`

So the value of `n` we are looking for is roughly `sqrt(4016) ~~ 63` and we find:

`T_63 = 1/2*63*64 = 2016`

`T_62 = 1/2*62*63 = 1953`

So `2008` lies on the reverse diagonal which has `T_63 = 2016` at one end.

Note that due to the boustrophedonic (like an ox ploughing a field) way in which the numbers are written, `T_63` will be written on the top row as the `63`rd term of that row.

`2008` will occur `8` rows below and to the left of it, i.e. on the `9`th row in the `55`th column.